Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits

Marks: 10M

Given:

$r_d = 40k\Omega$

$kn =0.4×10^{-3}$

$V_{GS(th)}=3V$

$Av=-g_m(r_d|| R_D)$

Rin=R1||R2

$Ro=r_d||R_D$

We know that,

$g_m= 2kn(V_{GSQ}-V_T)$

$g_m= 2×0.4×10^{-3}(V_{GSQ}-3)$……………..(1)

To find $g_m and V_{GSQ}$ we have to perform DC analysis, all the coupling capacitor acts as open circuit, as shown in Figure 2:

$I_{DS}=kn(V_{GS}-V_T)^2$

$I_{DS}=0.4\times10^{-3}(V_{GS}-3)^2$----------------(2)

But voltage divider biasing equation is,

$V_{R2}-V_{GS}-(I_{DS}\times R_S)=0$

$V_{GS}=V_{R2}-(I_{DS}\times1.2K\Omega)$----------------(3)

Using Voltage divider rule,

$V_{R2}=V_{GS}=\frac {V_{DD}\times{R_2}}{R1+R2}$

=$\frac {30\times10M\Omega}{10M\Omega+40M\Omega}$

=6V

Substitute in equation (3),

$V_{GS}=6-{(I_{DS}\times1.2K\Omega)}$------------------(4)

Put equation (4) in (2),

$I_{DS}=0.4\times10^{-3}(6-{(I_{DS}\times1.2K\Omega)}-3)^2$

$I_{DS}=0.4\times10^{-3}(3-{I_{DS}\times1.2K\Omega})^2$

Solving above equation we get two values of $I_{DS}$,

$I_{DS}=5.625mA or I_{DS}$=1.11mA

Hence from above observation, select $I_{DSQ}=1.11mA because V_{GS} \gt V_T$

Substitute in equation (4),

$V_{GSQ}=4.668V$

Hence substitute $V_{GSQ}$ in equation (1),

$g_m=2\times0.4\times10^{-3}(4.668-3)$

$g_m=1.3344\times10^{-3}$

$Av=-g_m(r_d||R_D)$

$Av=-1.3344\times10^{-3}(40k\Omega||3.3K\Omega)$

Av=-4.06

Rin=R1||R2

$Rin=40M\Omega||10M\Omega$

$Rin=8M\Omega$

$Ro=r_d||R_D$

$Ro=40M\Omega||3.3K\Omega$

$Ro=3.048K\Omega$